NCERT Solutions for class 8 Maths chapter 4 Data handling exercise 6.1

Exercise 6.1

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Introduction - Squares and Square Roots, Chapter 6 - NCERT Class 8th Maths Solutions

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Q 1 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 1
What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272
(iii) 799 (iv) 3853
(v) 1234 (vi) 26387
(vii) 52698 (viii) 99880
(ix) 12796 (x) 55555
Sol :
We know that if a number has its unit’s place digit as a, then its square will end with the unit digit of the multiplication a × a.
(i) 81
Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.
(ii) 272
Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.
(iii) 799
Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.
(iv) 3853
Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.
(v) 1234
Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.
(vi) 26387
Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.
(vii) 52698
Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.
(viii) 99880
Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.
(xi) 12796
Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.
(x) 55555
Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.

Q 2 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 2
The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453
(iii) 7928 (iv) 222222
(v) 64000 (vi) 89722
(vii) 222000 (viii) 505050
Sol :
The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeros at the end of it.
(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.
(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.
(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.
(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.
(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(vii) 222000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.
(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.

Q 3 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 3
(i) 431 (ii) 2826
(iii) 7779 (iv) 82004
Sol :
The square of an odd number is odd and the square of an even number is even. Here, 431 and 7779 are odd numbers.
Thus, the square of 431 and 7779 will be an odd number.

Q 4 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 4
Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1…2…1
10000001² = …
Sol :
In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,
100001² = 10000200001
10000001² = 100000020000001

Q 5 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 5
Observe the following pattern and supply the missing number.
11² = 121
101² = 10201
10101² = 102030201
1010101² = …
…² = 10203040504030201
Sol :
By following the given pattern, we obtain
1010101² = 1020304030201
101010101² = 10203040504030201

Q 6 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 6
Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + _ ² = 21²
5² + _ ² + 30² = 31²
6² + 7² + _ ² = __²
Sol :
From the given pattern, it can be observed that,
(i) The third number is the product of the first two numbers.
(ii) The fourth number can be obtained by adding 1 to the third number.
Thus, the missing numbers in the pattern will be as follows.
4² + 5² + = 21²
5² + + 30² = 31²
6² + 7² + =

Q 7 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 7
Without adding find the sum
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Sol :
We know that the sum of first n odd natural numbers is n².
(i) Here, we have to find the sum of first five odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = (5)² = 25
(ii) Here, we have to find the sum of first ten odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100
(iii) Here, we have to find the sum of first twelve odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)² = 144

Q 8 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 8
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11odd numbers.
Sol :
We know that the sum of first n odd natural numbers is n².
(i) 49 = (7)²
Therefore, 49 is the sum of first 7 odd natural numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11)²
Therefore, 121 is the sum of first 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q 9 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

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Question 9
How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Sol :
We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).
(i) Between 12² and 13², there will be 2 × 12 = 24 numbers
(ii) Between 25² and 26², there will be 2 × 25 = 50 numbers
(iii) Between 99² and 100², there will be 2 × 99 = 198 numbers